Problem: Is the function given below continuous/differentiable at $x=-4$ ? $g(x)=\begin{cases} \dfrac{1}{2}x^2&,&x<-4 \\\\ x^2+5x&,&x\geq-4 \end{cases}$ Choose 1 answer: Choose 1 answer: (Choice A) A Continuous but not differentiable (Choice B) B Differentiable but not continuous (Choice C) C Both continuous and differentiable (Choice D) D Neither continuous nor differentiable
Solution: Checking for continuity at $x=-4$ For the function to be continuous at $x=-4$, we need the two-sided limit $\lim_{x\to -4}g(x)$ to exist and be equal to $g(-4)$. This is the same as requiring that the two one-sided limits $\lim_{x\to -4^-}g(x)$ and $\lim_{x\to -4^+}g(x)$ exist and are equal to $g(-4)$. According to $g$ 's definition, $g(-4)=(-4)^2+5(-4)=-4$. $\lim_{x\to -4^-}g(x)$ : $\dfrac{1}{2}x^2$ evaluated at $x=-4$ is equal to $8$. Since $\dfrac{1}{2}x^2$ is continuous, we can be certain that $\lim_{x\to -4^-}g(x)=8$. $\lim_{x\to -4^+}g(x)$ : $x^2+5x$ evaluated at $x=-4$ is equal to $-4$. Since $x^2+5x$ is continuous, we can be certain that $\lim_{x\to -4^+}g(x)=-4$. The two limits exits, but they are not equal. Therefore, the function is not continuous at $x=-4$. Graphically, the function skips a step at this point. [I would like to see that, please!] Checking for differentiability at $x=-4$ Since the function isn't continuous at $x=-4$, it cannot be differentiable at that point. In conclusion, the function is neither continuous nor differentiable at $x=-4$.